Problem 6 - Calculus

Question

Find the derivatives of the functions in \(1-26\). $$ (x-1)^{2}(x-2)^{2} $$

Step-by-Step Solution

Answer

The derivative is \(4x^3 - 12x^2 + 12x - 4\).

Step 1: Identify the Function

The given function is \((x-1)^2(x-2)^2\). This is a product of two functions, \(u(x) = (x-1)^2\) and \(v(x) = (x-2)^2\). To differentiate, we will use the product rule and the chain rule.

Step 2: Apply the Product Rule

The product rule states if you have \(u(x)\cdot v(x)\), then the derivative is \(u'(x) v(x) + u(x) v'(x)\). So, first, we need to find \(u'(x)\) and \(v'(x)\).

Step 3: Differentiate \((x-1)^2\)

Using the chain rule, the derivative of \(u(x) = (x-1)^2\) is \(u'(x) = 2(x-1)\cdot 1\). This simplifies to \(u'(x) = 2(x-1)\).

Step 4: Differentiate \((x-2)^2\)

Similarly, using the chain rule, the derivative of \(v(x) = (x-2)^2\) is \(v'(x) = 2(x-2)\cdot 1\). This simplifies to \(v'(x) = 2(x-2)\).

Step 5: Insert Derivatives into Product Rule

Now substitute the derivatives into the product rule: \(u'(x)v(x) + u(x)v'(x) = [2(x-1)(x-2)^2] + [(x-1)^2 2(x-2)]\).

Step 6: Simplify the Expression

To simplify, distribute and combine like terms for \(2(x-1)(x-2)^2 + 2(x-2)(x-1)^2\). This results in: 2(x-1)(x^2 - 4x + 4) + 2(x-2)(x^2 - 2x + 1).Expanding both terms and combining them gives: 2(x^3 - 4x^2 + 4x - x^2 + 3x - 1 + x^3 - 2x^2 + x - 2x^3 + 4x^2 - 2) = 4x^3 - 12x^2 + 12x - 4.

Step 7: Finalize the Derivative

Focusing on the terms, collect and simplify them to get the final derivative, which is:\(\frac{d}{dx}[(x-1)^2(x-2)^2] = 4x^3 - 12x^2 + 12x - 4\).

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of a product of two functions. In the case of the function \((x-1)^2(x-2)^2\), we can think of it as consisting of two components: \(u(x) = (x-1)^2\) and \(v(x) = (x-2)^2\). The product rule tells us that the derivative of a product \(u(x) \cdot v(x)\) is given by

• \(u'(x) \cdot v(x) + u(x) \cdot v'(x)\).

To apply this rule, we first find the derivatives \(u'(x)\) and \(v'(x)\). Each of these derivatives requires additional tools like the chain rule to compute, and once calculated, they can be plugged back into the product rule formula to find the derivative of the entire expression. Understanding and applying the product rule correctly will help you navigate more complex derivative problems effectively.
Remember, the essence of the product rule is to allow us to differentiate products layer by layer, maintaining the structure of each separate function while considering their interaction with each other.

Chain Rule

The chain rule is an indispensable technique for differentiation when dealing with composite functions. A composite function is one where one function nests inside another. For example, in the expression \((x-1)^2\) or \((x-2)^2\), we are looking at something like \((g(x))^2\) where the inner function \(g(x)\) is \(x-1\) or \(x-2\) respectively. Use the chain rule when you need to differentiate a function of a function.
The chain rule states that the derivative of \((g(x))^n\) is

• \(n(g(x))^{n-1} \cdot g'(x)\).

In our original exercise, the chain rule was used in differentiating each part of the function:

• For \((x-1)^2\), the derivative is \(2(x-1)\cdot 1\), simply written as \(2(x-1)\).
• For \((x-2)^2\), the process is identical, yielding \(2(x-2)\).

Each of these results was part of the steps needed to fully utilize the product rule as described previously. The chain rule is often used in combination with other rules like the product rule, allowing you to break down even the most daunting functions into simpler, more manageable pieces.

Polynomial Differentiation

Polynomial differentiation is the process of finding the derivative of polynomial expressions. Polynomials are algebraic expressions consisting of variables and coefficients exhibiting non-negative integer exponents. Differentiating polynomials is a straightforward process, once you know the rules: each term's exponent is multiplied by the coefficient, and then reduced by one.
Consider a simple function \(f(x) = ax^n\). The derivative is found by

• Multiplying the term's coefficient \(a\) by its power \(n\),
• And reducing the power by one, i.e., \(anx^{n-1}\).

For the polynomial resulting from the multiplication of our differentiated components, like \(4x^3 - 12x^2 + 12x - 4\), we apply these principles. Each term is treated independently before aggregating them to present the complete derivative.
In addition to basic rules, polynomial differentiation often involves the application of other derivative rules, such as the chain rule and the product rule, especially when dealing with polynomials within larger and more complex expressions. By understanding polynomial differentiation, you create a solid foundation to tackle various calculus problems efficiently.